Problem 107 - Chemistry: The Central Science

Question

Barium azide is \(62.04 \%\) Ba and \(37.96 \%\) N. Each azide ion has a net charge of \(1-\). (a) Determine the chemical formula of the azide ion. (b) Write three resonance structures for the azide ion. (c) Which structure is most important? (d) Predict the bond lengths in the ion.

Step-by-Step Solution

Answer

The chemical formula of the azide ion is N₆³⁻. Three resonance structures are essential in representing the azide ion, with the first and second resonance structures being the most important due to the least separation of charge. The predicted bond lengths in the azide ion are approximately 135 pm.

Step 1: Calculate the molar mass of Ba and N

First, we need to determine the number of moles of Ba and N in the compound. To do this, we will use the molar masses of Ba and N, which are 137.33 g/mol and 14.01 g/mol, respectively. Using the given percentage composition, we can calculate the moles of Ba and N. For Ba: Percentage mass of Ba = 62.04% Let's assume we have 100 g of Barium azide, then the mass of Ba would be 62.04 g. Now we can find the moles of Ba: moles of Ba = mass of Ba / molar mass of Ba moles of Ba = 62.04 g / 137.33 g/mol moles of Ba ≈ 0.452 For N: Percentage mass of N = 37.96% In a 100 g sample of Barium azide, the mass of N would be 37.96 g. Now we can find the moles of N: moles of N = mass of N / molar mass of N moles of N = 37.96 g / 14.01 g/mol moles of N ≈ 2.710 Now we find the ratio between the moles of Ba and N: Mole ratio Ba:N ≈ 0.452:2.710 To find the simplest ratio, we divide both sides by the smallest number (0.452): Ba:N ≈ 1:6 So, the formula for the azide ion is N₆³⁻ (since the charge on azide ion is 1-).

Step 2: Write the Lewis structure for the azide ion

Now that we have the formula for the azide ion (N₆³⁻), we can move forward to writing the Lewis structure. Since we are looking for a resonance structure, we will focus on the nitrogen atoms. Each nitrogen atom has 5 electrons in its valence shell, and overall, the azide ion has a negative charge. Therefore, there will be a total of 5*6 + 1 = 31 electrons that need to be distributed. First, we arrange the nitrogen atoms linearly, in a way that equally distributes the valence electrons between them. We put single bonds between the nitrogen atoms, which will take 4 electrons. The remaining 31 - 4 = 27 electrons are distributed among the nitrogen atoms so that they satisfy the octet rule (8 electrons around each atom).

Step 3: Draw three possible resonance structures for the azide ion

Now that we have the basic Lewis structure, we can move on to drawing resonance structures. Resonance structures are different configurations of the same molecule that have the same overall distribution of electrons. In this case, we can draw three resonance structures as follows: 1. First resonance: Start with a single bond between N1 and N2, a double bond between N2 and N3, and the remaining electrons are distributed as lone pairs. 2. Second resonance: Start with a double bond between N1 and N2, a single bond between N2 and N3, and the remaining electrons are distributed as lone pairs. 3. Third resonance: Start with a double bond between N1 and N2, a double bond between N2 and N3, an additional electron on N1 as a lone pair, and the remaining electrons are distributed as lone pairs.

Step 4: Determine the most important resonance structure

To find the most important resonance structure, we need to consider the formal charges on each atom and look for a structure with the least separation of charge. In the first resonance, the formal charge on N1 = 0, N2 = 0, and N3 = -1. In the second resonance, the formal charge on N1 = -1, N2 = 0, and N3 = 0. In the third resonance, the formal charge on N1 = -1, N2 = 1, and N3 = -1. Since the structures with evenly distributed charges are more favorable, the most important resonance structure would be the first or the second resonance, as they have the least separation of charge.

Step 5: Predict the bond lengths in the azide ion

The bond lengths in the azide ion can be predicted based on the most important resonance structure(s). Since resonance structures 1 and 2 are equally important, the actual bond lengths in the azide ion will be an average of the single and double bond lengths. A single N-N bond has a bond length of around 145 pm (picometers), while a double N-N bond has a bond length of around 125 pm. The actual bond lengths in the azide ion will be an average of these two values: Average bond length ≈ (145 pm + 125 pm) / 2 ≈ 135 pm So, the bond lengths in the azide ion are expected to be around 135 pm.